Coefficient of kinetic friction, need today

[an9890]

coefficient of kinetic friction, URGENT need today

The system shown in the figure below has an acceleration of magnitude 0.51 m/s2, where m1 =
4.30 kg and m2 = 6.00 kg. Assume that the coefficient of kinetic
friction between block and incline is the same for both inclines.

Link to figure:
http://img9.imageshack.us/img9/3437/figureyh.jpg
m1 is box on the left, m2 is box on the right. both angles are 35 degrees. a) find the coefficient of kinetic friction

Here's what I tried:
coefficient of kinetic friction = (gsinΘ - a) / gcos Θ
and got .63668

when i input my answer in the system said "your answer differs from the correct answer by orders of magnitude."

b) find the tension in the string
T= m1a + m1gsinΘ
= 26.36 N

system said "your answer differs by 10% of correct answer..."Please help. I need this by tonight. Thank you!

 

[hotvette]

Let's start with the basics. What equation of motion (i.e. \SigmaF = ma) did you use for each mass?

 

[an9890]

^

(a)
∑ Fx = ma --> mgsinΘ - f_k = ma
∑ Fy = 0 --> n - mgcosΘ = 0
n = mgcosΘ
and f_k = mu_kn

so mu_k = (gsinΘ -a) / gcos Θ

 

[hotvette]

∑ Fx = ma --> mgsinΘ - fk = ma

I presume that's for m2. But, you forgot the force of the tension. Also, you need a similar equation for m1. You'll then have two equations in two unknowns (T and mu) that can be solved by substitution.